A block of mass 1kg is suspended freely by a thread. A bullet of mass 3kg is fired horizontally at the block and it becomes embedded in it. The block swings to one side using a vertical distance of 50cm. With what speed did the bullet hit the block?
Given:
"m=1\\:\\rm kg"
"M=3\\:\\rm kg"
"h=0.5\\:\\rm m"
Let the initial speed of a bullet is denoted as "u". The law of conservation of momentum gives
"Mu=(m+M)v"The law of conservation of energy gives
"\\frac{(M+m)v^2}{2}=(M+m)gh"Hence, we get
"u=\\frac{M+m}{M}v=\\frac{M+m}{M}\\sqrt{2gh}""u=\\frac{3+1}{3}\\sqrt{2*9.8*0.5}=4.17\\:\\rm m\/s"
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